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Let $S= \{0, 1, 2, 3, 4, 5, 6, 7\}$ and $⊗$ denote multiplication modulo $8,$ that is, $x ⊗ y= (xy) \mod 8$

1. Prove that $( \{ 0, 1\}, ⊗)$ is not a group.
2. Write three distinct groups $(G, ⊗)$ where $G ⊂ S$ and $G$ has $2$ elements.

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$\begin{array}{c|cc} ⊗&0&1\\ \hline 0&0&0\\ 1&0&1 \end{array}$

a. $1$ is the identity element. Inverse does not exist for zero. So, it is not a group.

b. $\begin{array}{c|cc} ⊗&1&3\\ \hline 1&1&3\\ 3&3&1 \end{array}\qquad\begin{array}{c|cc} ⊗&1&5\\\hline1&1&5\\ 5&5&1 \end{array}\qquad\begin{array}{c|cc} ⊗&1&7\\\hline1&1&7\\ ​7&7&1 \end{array}$

by

Then how can A be a group?
Sorry I can't get your question.Actual question is to prove A is not a group.I also proved it.Is there is any mistake?
Sorry. i misread the question and didn't see the A, B separation in answer.
it's ok :)

({0,1},⊗) is not a group  because : (0e)mod8=0 is haing e=0 as identity but (1e)mod8=1 should have 1 as identity so no single identity is there.

({1,7}, ⊗ ) :

1. closed

2. associative

3. Identity = 1

4. inerse:  1-1=7 , 7-1=1

can someone suggest 2 more subgroups.

{1,3,5,7} with order 4

with order 2 no further subgroup exist
Inverse of 7 is 7

The other two groups are {1,3}, {1,5}